JOHN's DECOMPOSITION of the IDENTITY in the NON-CONVEX CASE Jesus Bastero* and Miguel Romance** 1. Introduction and Notation T

JOHN’S DECOMPOSITION OF THE IDENTITY IN THE NON-CONVEX CASE

Jesus Bastero* and Miguel Romance**

Abstract We prove an extension of the classical John’s Theorem, that characterices the ellipsoid of maximal volume position inside a convex body by the existence of some kind of decomposition of the identity, obtaining some results for maximal volume position of a compact and connected set inside a convex set with nonempty interior. By using those results we give some estimates for the outer volume ratio of bodies not necesarily convex.

1. Introduction and Notation n Throughout this paper, we consider R with the canonical basis (e1, . . . , en) and its usual Euclidean structure , . Let Bn = x Rn ; x = x, x 1/2 1 be the euclidean ball h· ·i 2 { ∈ | | h i ≤ } on Rn. If K Rn, then int K, Kc and ∂K will denote the interior, the complementary and the border⊆ of K, respectively; conv(K) will be the convex hull of K, K0 will denote the polar of K with respect to the origin, i.e. K0 = y Rn ; x, y 1, x K and { ∈ h i ≤ ∀ ∈ } vol (K) represents the Lebesgue measure on Rn of K. n n n Following [TJ], if K1 K2 R , we call a pair (x, y) R R a contact pair for (K ,K ) if it satisfies: i) x⊆ ∂K⊆ ∂K , ii) y ∂K0 and∈ iii) x,× y = 1. 1 2 ∈ 1 ∩ 2 ∈ 2 h i As it is usual y x denotes the linear transformation on Rn defined by y x(z) = ⊗ n ⊗ z, y x and In will be the identity map on R . h i John’s ellipsoid theorem is a classical tool in the theory of convex bodies; it says how far a convex body is from being an ellipsoid. John showed that each convex body contains a unique ellipsoid of maximal volume and characterized it. The decomposition of the identity associated to this characterization gives an effective method to introduce an appropiated euclidean structure in finite dimensional normed spaces, when we consider centrally symmetric convex bodies. We can state John’s theorem in the following way:

Theorem ([J], [Ba2], [Ba3]). Let K be a convex body in Rn and suppose that the n euclidean ball B2 is contained in K, then the following assertions are equivalent: n (i) B2 is the ellipsoid of maximal volume contained in K, n (ii) there exist λ1, . . . , λm > 0 and u1, . . . , um ∂K ∂B2 , with m n(n + 3)/2 such m m ∈ ∩ ≤ that I = u u and λ u = 0. n i ⊗ i i i i=1 i=1 n X X (iii) B2 is the unique ellipsoid of maximal volume contained in K.

* Partially supported by Grant D.G.E.S.I.C. (Spain) ** Supported by F.P.I. Grant D.G.E.S.I.C. (Spain)

1 One can consider this situation for any general couple of convex bodies (K1,K2) n n or, even more, for any couple of compact sets in R instead of (B2 ,K). Suppose that n n K1 is a compact set in R with vol (K1) > 0 and K2 is another compact set in R , with int K2 = . A compactness argument shows that there exists an aﬃne position of K1, namely6 K˜∅ , such that K˜ K and 1 1 ⊆ 2 ˜ n vol (K1) = max vol (a + T (K1)); a + T (K1) K2, a R ,T GL(n) . { ⊆ ∈ ∈ }

This position K˜1 is called maximal volume position of K1 inside K2. Very re- cently, Giannopoulos, Perisinaki and Tsolomitis have considered the convex situation and proved the following n Theorem [G-P-T]. Let K1 K2 be two smooth enough convex bodies in R such that ⊆ K1 is in maximal volume position inside K2. Then for every point z in the interior of K1, 2 there exist λ1, . . . , λN > 0, with N n + n + 1 and contact pairs for (K1 z, K2 z), ≤ N N− − (x z, y ),(1 i N) such that: (i) λ y = 0 and (ii) I = λ y x . i − i ≤ ≤ i i n i i ⊗ i i=1 i=1 X X Furthermore, if we assume the extra assumption for K1 to be a polytope and K2 to have (2 boundary with strictly positive curvature, then a center z can be chosen in C 1 N K vertices of K for which we have (i), (ii) and also (iii) λ x = z. 1 \{ 1} n i i i=1 X The special case of considering K1 and K2 centrally symmetric convex bodies was first observed by Milman (see [TJ], Theorem 14.5). The aim of this paper is to extend this result to the non-convex case. We obtain a n general result which is valid for K1 a compact, connected set in R with vol (K1) > 0, n K1 K2, where K2 is a compact in R such that int conv(K2) = and K1 is in maximal ⊆ 6 ∅ volume position inside conv(K2) (no extra assumptions on the boundary of the bodies are used). The method we develop to prove our result is different from that in [G-P-T]. We follow the ideas given in [Ba3], with suitable modifications and the main result we achieve is the following n Theorem 1.1. Let K1 R be a connected, compact set with vol (K1) > 0 and n ⊆ K2 R be a compact set such that K1 K2. If K1 is in maximal volume position ⊆ ⊆ 2 inside conv(K2), for every z int conv(K2) there exist N N, N n + n, (xi, yi) contact pairs for (K z, K ∈ z) and λ > 0 for all i = 1,...,N∈ such≤ that: 1 − 2 − i N 1 λ y x = I k k ⊗ k n n kX=1 N

λk yk = 0. kX=1 It is well known that if there exists a decomposition of the identity in the sense of theorem 1.1 we can not expect that K1 were the unique maximal volume position inside

2 K2 even for convex bodies, as it can be shown by considering simpleces or octahedra inscribed in the cube. Furthermore, an equivalence as it appears in John’s Theorem is not true in general. We study this problem and as a consequence we obtain n Theorem 1.2. Let K1 R be a connected, compact set with vol (K1) > 0 and n ⊆ K2 R be a compact set such that K1 K2. Let z be a ﬁxed point in int conv (K2). Then⊆ the following asumptions are equivalents:⊆ n (i) vol (K1) = max vol (a + S(K1)); a R , a + S(K1) conv(K2) , where S runs over all symmetric{ positive deﬁnite matrices.∈ ⊆ } n2+3n (ii) There exist N N, N , (xi, yi) contact pairs for (K1 z, K2 z) and ∈ ≤ 2 − − λi > 0 for all i = 1,...,N such that:

N 1 λ (y x + x y ) = I (1) k k ⊗ k k ⊗ k n n kX=1 N

λkyk = 0. kX=1 (iii) K1 is the unique position of K1 verifying (i). In section 3 we extend the upper estimates of the volume ratio proved in [G-P-T] by deﬁning the outer volume ratio of a compact K1 with respect to a convex body K2, by considering an appropriate index. We follow the methods that appears in [G-P-T] by using Brascamp-Lieb and reverse Brascamp-Lieb inequalities as the main tools.

2. Proofs of Main Theorems n Throughout this section K1 will be a connected, compact set in R with vol (K1) > 0 n and K2 will be a compact in R such that K1 K2. Therorem 1.1 gives us a suﬃcient condition for the existence of some kind of John’s⊆ decomposition of the identity and it follows the spirit of the work of K.M. Ball (see for instance [Ba2] or [Ba3]). Proof of Theorem 1.1: First of all, notice that int conv (K ) = , since K 2 6 ∅ 1 ⊆ conv (K2) and vol (K1) > 0. Without loss of generality we can assume that z = 0. Furthermore, since K K and conv(K )0 = K0 a contact pair for (K , conv(K )) is 1 ⊆ 2 2 2 1 2 also a contact pair for (K1,K2), so, we may suppose conv(K2) = K2. n n n Let = (y x, y) (R , R ) R ;(x, y) is a contact pair for (K1,K2) . By using the maximalityA { of⊗ the volume∈ L of K×, the convexity of K and since 0 }int K it is 1 2 ∈ 2 easy to prove that is a non empty subset of (Rn, Rn) Rn. We will show that 1 A L × n n n ( n In, 0) conv( ), where conv( ) is the convex hull of in (R , R ) R . Suppose∈ thatA ( 1 I , 0) / conv(A ). Then by using a separationA L theorem,× there exist n n ∈ A H (Rn, Rn) and b Rn such that: ∈ L ∈ 1 I ,H + 0, b > y x, H + b, y hn n itr h i h ⊗ itr h i

n n for all (x, y) contact pair and where , tr denotes the trace duality on (R , R ), i.e. T,S = tr ST . h· ·i L h itr 3 Thus for every contact pair (x, y) 1 tr H > Hx, y + b, y . (2) n h i h i

There is no loss of generality to assume tr H = 0. Indeed, we can choose H˜ = H tr H n n − In (R , R ), which is a linear operator with trace zero that veriﬁes: n ∈ L tr H tr H˜ Hx,˜ y + b, y = Hx, y x, y + b, y < 0 = h i h i h i − n h i h i n

for all (x, y) contact pair. By using the linear map defined by the matrix H and b Rn ∈ we are going to construct a family of affine maps Sδ’s, with 0 < δ < δ1, such that det S 1 and S (K ) int K , which contradicts the maximality of the volume of | δ| ≥ δ 1 ⊆ 2 K1. We will divide the proof of that fact into 3 steps. By continuity, there exists a positive number δ0 > 0 such that In δH is invertible n n − for all 0 < δ < δ0. For each 0 < δ < δ0 we take Sδ : R R defined by Sδ(z) = 1 1 −→ (I δH)− (z) + δ(I δH)− (b). n − n − Step 1: There exists 0 < δ1 δ0 such that Sδ(K1) ∂K2 = , for all 0 < δ < δ1. Consider ≤ ∩ ∅

M = x ∂K ; y K0 such that x, y = 1 and Hx, y + b, y 0 . ∈ 2 ∃ ∈ 2 h i h i h i ≥ It is easy to check that M is a compact subset of ∂K2 and also M ∂K1 = . If there 0 ∩ ∅ exists an x M ∂K1 ∂K2 ∂K1, there would exist y K2 such that x, y = 1 (so (x, y) is∈ a contact∩ pair)⊆ and ∩Hx, y + b, y 0 = tr H which∈ would contradicth i (2). c h i h i ≥ Therefore M K1; by compactness of M and by continuity, there exists 0 < δ1 ( δ0) such that (I ⊆ δH)(M) δb Kc, for all 0 < δ < δ , and so S (K ) M = . ≤ n − − ⊆ 1 1 δ 1 ∩ ∅ Now let x ∂K2. We will prove that x / Sδ(K1). We only have to consider the case x / M. Then∈ ∈ ∈ Hx, y + b, y < 0 h i h i 0 for all y K2 such that x, y = 1. Since 0 int K2 and K2 is a convex body there exists y ∈ K0 such that hx, y i = 1, so we have∈ that 0 ∈ 2 h 0i x δHx δb, y = 1 δ ( Hx, y + b, y ) > 1 h − − 0i − h 0i h 0i

for all δ > 0 and in particular for all 0 < δ < δ1. Hence x δHx δb / K1, or equivalently x / S (K ). − − ∈ ∈ δ 1 Step 2: For every 0 < δ < δ there exists λ > 1 such that λ S (K ) int K . 1 δ δ δ 1 ⊆ 2 Note that Sδ(K1) is connected and Sδ(K1) ∂K2 = , therefore either Sδ(K1) int K or S (K ) Kc. Fix x K int K , it exists∩ since∅ vol (K ) = 0, and take ⊆ 2 δ 1 ⊆ 2 ∈ 1 ∩ 2 1 6 C = S (x) ; 0 δ < δ . x { δ ≤ 1}

It is easy to check that Cx is connected, Cx ∂K2 = , Cx int K2 = and Cx int K . Therefore S (K ) int K = and by∩ connectedness∅ ∩ of K we6 conclude∅ that⊆ 2 δ 1 ∩ 2 6 ∅ 1 4 S (K ) int K , for all 0 < δ < δ . Now, by a compactness argument and the fact that δ 1 ⊆ 2 1 Sδ(K1) int K2 we conclude that for every 0 < δ < δ1 there exists λδ > 1 such that λ S (K⊆) int K . δ δ 1 ⊆ 2 Step 3: vol (λδ Sδ(K1)) > vol (K1), for all 0 < δ < δ1 . Indeed, λn vol (K ) vol (λ S (K )) = δ 1 . δ δ 1 det(I δH) | n − | Now, by using the inequality between arithmetic mean and geometric mean (denoted 1 n 1 brieﬂy AM-GM inequality) we obtain that det(In δH) n tr (In δH) = 1, and so | − | ≤ − vol (λ S (K )) λn vol (K ) > vol (K ). δ δ 1 ≥ δ 1 1 1 Therefore, if ( n In, 0) / conv , then K1 is not in maximal volume position inside K2. Note that the fact∈ that AN n2 + n is deduced from the classical Caratheodory’s 1 ≤ 2 theorem since (y x n In, y);(x, y) is a contact pair is contained in a (n + n 1) dimensional vector{ ⊗ space.− } −

Remarks. 1) For every K Rn with vol (K) > 0 it is easy to check that K is in maximal volume position inside⊆ conv (K). 2) We note that K K and K is in maximal volume position inside conv (K ) 1 ⊆ 2 1 2 implies that K1 is in maximal volume position inside K2. The converse is not true, as the following example shows. Consider

n K1 = x R ; x = max xi 1 { ∈ k k∞ 1 i n | | ≤ } ≤ ≤ K = K 2∂K . 2 1 ∪ 1

It is trivial to see that K1 is in maximal volume position inside K2, K1 is not in maximal volume position inside conv(K2) and there is no decomposition of the identity as before, since there are no contact pairs for (K1,K2). n Corolary 2.1. Let K1 K2 R be as in the theorem 1.1. If conv (K1) is a polytope, (2) ⊆ ⊆ conv (K2) has boundary with strictly positive curvature and K1 is maximal volume C 2 position inside conv (K2), then there exist z conv (K1), N N, N n + n, (xk, yk) contact pairs of (K z, K z) and λ > 0∈for all k = 1,...,N∈ such≤ that 1 − 2 − k N 1 λ y x = I k k ⊗ k n n kX=1 N N

λk yk = λk xk = 0. kX=1 kX=1 Proof: It is easy to prove that the fact that K1 is in maximal volume position inside conv (K2) implies that conv (K1) is in maximal volume position inside conv (K2) and ∂K ∂(conv (K )) = ∂(conv (K )) ∂(conv (K )). Therefore we can assume that 1 ∩ 2 1 ∩ 2 5 (2) K1 is a polytope and K2 is a convex which has a boundary with strictly positive curvature. But notice that this case was studied byC A. Giannopoulos, I. Perissinaki and A. Tsolomitis (see [G-P-T]) concluding the result needed.

Now we can ask if the existence of some kind of decomposition of the identity in Rn would imply that K1 were the unique maximal volume position inside K2, as it happens in the classical John’s Theorem. It is well known that we can’t expect such a thing, simply by considering simplices or octahedra inscribed in the cube. Theorem 1.2 shows, loosely speaking, that not only the existence of a “modified” John’s decomposition of the identity for a pair (K1,K2) implies that K1 is the unique “pseudo” maximal volume position inside K2, but also that this “pseudo” maximality implies the existence of a “modified” decomposition of the identity too. Proof of Theorem 1.2: As before, we can show that int conv (K ) = . We can also 2 6 ∅ assume z = 0 and K2 convex. 1 n n n (i) (ii) Let = ( 2 (y x + x y), y) (R , R ) R ;(x, y) is a contact pair . By using⇒ the maximalityB { of the⊗ volume⊗ of K ∈, the L convexity× of K and since 0 int K} 1 2 ∈ 2 it is easy to prove that is a non empty subset of (Rn, Rn) Rn. As in the proof of B 1 L × theorem 1.1, we will show that ( n In, 0) conv ( ). Suppose, on the contrary, that ( 1 I∈, 0) / Bconv ( ). Then by using a separation n n ∈ B theorem, there exist H (Rn, Rn) and θ Rn such that: ∈ L ∈ 1 1 I ,H + 0, θ > ( y x, H + x y, H ) + θ, y hn n itr h i 2 h ⊗ itr h ⊗ itr h i for all (x, y) contact pair. Therefore 1 1 tr H > ( Hx, y + x, Hy ) + θ, y . n 2 h i h i h i There is no loss of generality to assume that: ˜ 1 ? 1) H is a symmetric matrix because in other case we could take H = 2 (H + H ) which is a symmetric matrix that verifies that Hx,˜ y + x, Hy˜ = Hx, y + x, Hy and therefore for every contact pair (x, y) h i h i h i h i 1 tr H > Hx, y + θ, y . n h i h i ˜ tr H n n 2) tr H = 0 since in other case we can choose H = H n In (R , R ) which is a linear operator with trace zero that verifies: − ∈ L

tr H tr H˜ Hx,˜ y + θ, y = Hx, y x, y + θ, y < 0 = h i h i h i − n h i h i n for all (x, y) contact pair. Therefore there exist a symmetric matrix H (Rn, Rn) with tr H = 0 and θ Rn such that: ∈ L ∈ 0 > Hx, y + θ, y h i h i 6 for all (x, y) contact pair. By using the linear map defined by the matrix H and θ Rn ∈ we are going to construct a family of affine maps Tδ( ) = Sδ( ) + bδ, with Sδ symmetric positive definite matrix for all 0 < δ < δ , such that · det S · 1 and T (K ) int K , 1 | δ| ≥ δ 1 ⊆ 2 which contradicts the maximality of K1. By continuity, there exists a positive number δ > 0 such that I δH is invertible 0 n − and symmetric positive definite for all 0 < δ < δ0. For each 0 < δ < δ0 we take n n 1 1 Tδ : R R defined by Tδ(z) = (In δH)− (z) + δ(In δH)− (θ). By the same methods−→ as in the proof of Theorem 1.1,− we can show that: − 1) There exists 0 < δ1 δ0 such that Tδ(K1) ∂K2 = , for all 0 < δ < δ1. 2) For every 0 < δ < δ≤there exists λ > 1 such∩ that λ∅ T (K ) int K . 1 δ δ δ 1 ⊆ 2 3) vol (λδ Tδ(K1)) > vol (K1), for all 0 < δ < δ1 which contradicts the maximality of K1. n (ii) (iii) Let T ( ) = S( ) + a be such that T (K1) K2, a R and S is a symmetric⇒ positive definite· matrix.· It is well known that we⊆ can find∈ an orthogonal matrix U O(n) and a diagonal matrix D with diagonal elements α1, . . . , αn > 0 such that S = U∈?DU and therefore

n ? vol (T (K1)) = det (U DU) vol (K1) = αk vol (K1). (3) | | ! kY=1

Hence we have to estimate αk. On the one hand, we obtain that

Q n U ?D Ux, y = α U ?e , x U ?e , y h i jh j ih j i j=1 X for all x, y Rn, by straightforward computation. On the∈ other hand, if (x, y) is a contact pair then T x, y 1 and therefore h i ≤ N N N 1 = λ λ T x , y = λ U ?D Ux , y k ≥ k h k ki k h k ki kX=1 kX=1 kX=1 N n n N = λ α U ?e , x U ?e , y = α λ U ?e , x U ?e , y k jh j kih j ki j k h j ki h j ki j=1 j=1 ! kX=1 X X kX=1 1 n 1 n = α U ?e ,U ?e = α . n jh j ji n j j=1 j=1 X X

1 Now by using the AM-GM inequality, we conclude that 1 1 α ( α ) n , which ≥ n j ≥ j implies that in (3) we obtain vol (T (K1)) vol (K1). ≤ P Q In addition to this, note that if T is such that vol (T (K1)) = vol (K1), then, by the equality case in the AM-GM inequality we would have that α1 = ... = αn = 1, so T = In + a. Therefore we would obtain that

1 T x, y = x + a, y = 1 + a, y ≥ h i h i h i 7 for all (x, y) contact pair and, in particular, a, yk 0 for all (xk, yk) contact pair that appears in the decomposition of the identity.h Buti we ≤ also would have that:

N N λ a, y = a, λ y = 0 kh ki h k ki kX=1 kX=1 which would imply that, a, y = 0 for all (x , y ) and then we would conclude that h ki k k 1 N a, a = λ a, y a, x = 0. n h i kh kih ki kX=1 Hence T =In.

Corolary 2.2. Let K1 K2 be as in theorem 1.1. Fix z int conv (K2). Then the following assumptions are⊆ equivalents: ∈ n (i) vol (K1) = max vol (a + S(K1)); a R , a + S(K1) conv(K2) , where S runs over all symmetric{ positive deﬁnite matrices.∈ ⊆ } (ii) For every S GL(n) symmetric matrix and every θ Rn there exists a contact pair (x, y) for∈(K z, K z) such that ∈ 1 − 2 − tr S Sx, y + θ, y . n ≤ h i h i Proof: As before, we can show that int conv (K ) = . We can also assume z = 0 and 2 6 ∅ K2 convex. (i) (ii) By Theorem 1.2 there exist (x , y ) contact pairs for (K z, K z) ⇒ i i 1 − 2 − and λi > 0 for all i = 1,...,N such that:

N N 1 λ (y x + x y ) = I and λ y = 0. k k ⊗ k k ⊗ k n n k k kX=1 kX=1 Suppose that there would exist S GL(n) symmetric matrix and θ Rn such that for every (x, y) contact pair ∈ ∈ tr S > Sx, y + θ, y . n h i h i Therefore

N N tr S tr S 1 = + θ, λ y = (S, θ), ( I , λ y ) = n n h i ii h n n k k i i=1 i=1 X X N N 1 = λ (S, θ), ( (y x + x y ), y ) = λ ( Sx , y + θ, y ) < ih 2 i ⊗ i i ⊗ i i i i h i ii h ki i=1 i=1 X X N tr S tr S < λ = i n n i=1 X 8 which leads us to a contradiction. (ii) (i) By using the hypothesis, for every H GL(n), θ Rn, there exists a contact pair⇒ such that ∈ ∈

1 1 tr H > ( Hx, y + x, Hy ) + θ, y n 2 h i h i h i

1 1 n n n which make that ( n , 0) conv ( ( 2 (y x + x y), y) (R , R ) R ; where (x, y) is a contact pair ) ∈ { ⊗ ⊗ ∈ L × }

Remarks. 1) If K1 is in maximal volume position inside conv (K2), then K1 is unique if we only consider aﬃne transformations given by symmetric, positive deﬁnite matrices. Indeed, 1 N 1 N this is due to the fact that n I = k=1 λk yk xk implies that n I = k=1 λk xk yk. n ⊗ n ⊗ 2) If we suppose either K1 =B or conv(K2)=B in the last theorem, we obtain a P2 2 P stronger conclusion, since the existence of contact pairs (xk, yk) and λk > 0 such that

N λ 1 N k (y x + x y ) = I and λ y = 0 2 k ⊗ k k ⊗ k n n k k kX=1 kX=1

is equivalent to the fact that vol (K1) = max vol (a + T (K1)) such that a + T (K1) n { ⊆ conv (K2), a R and T GL(n) and this maximum is only attained at K1, up ∈ ∈ } to orthogonal transformation (i.e. if vol (T (K1)) = vol (K1), then T is an orthogonal transformation). This is the classical John’s result. Let’s see it briefly. Suppose that there exists a decomposition of the identity (in the sense of (1)). If n n we take conv(K2) = B2 and T is an affine transformation such that T (K1) B2 , then there exist orthogonal matrices U, V , a diagonal matrix D with diagonal elements⊆ n ˜ α1, . . . , αn > 0 and a R such that T ( ) = VDU( ) + a. Now if we choose T ( ) = U ?DU( ) + (VU)?(a) then∈ it is easy to check· that this· map verifies: · · (a) U ?DU is a symmetric positive definite matrix. (b) T˜(K ) (VU)?(Bn) = Bn (since T˜( ) = (VU)?T ( )). 1 ⊆ 2 2 · · (c) vol (T˜(K1)) = vol (T (K1)).

Therefore by using (ii) (iii) in theorem 1.2 and since T˜ satisﬁes (a) and (b) we conclude that ⇒ vol (T (K )) = vol (T˜(K )) vol (K ) 1 1 ≤ 1 and the equality is only attained if T˜ = In, and so T is an orthogonal transformation. n Note that a similar reasoning can be applied to the case K1 = B2 .

9 3. Some estimates for the outer volume ratio of compact sets n n We can extend the notion of volume ratio to a pair (K1,K2) R R , where K2 is a ⊆ × convex body and K1 is a compact set with vol (K1) > 0, simply by n n Deﬁnition 3.1. Let K1 R be compact set with vol (K1) > 0 and K2 R be a convex body. We deﬁne outer⊆ volume ratio as ⊆

1 vol (K2) n vr(K2; K1) = inf 1 ; T aﬃne transformation with T (K1) K2 . (vol (T (K1)) n ⊆ )

It is quite easy to show that we cannot expect any upper estimate without asuming extra asumptions. We are going to introduce an index for compact sets with positive volume in order to get general bounds, depending only on the dimension and on the index, for the outer volume ratio with respect to a convex body. We recall that a set K Rn is p-convex, (0 < p 1) if λx + µy K, for every x, y K and for every λ, µ ⊆ 0 such that λp + µp = 1.≤ The p-convex hull∈ of a set K, which∈ we denote by p conv≥ (K), is defined as the intersection of all p-convex sets that contain K. It is easy to− see that 0 p conv (K). ∈ − Definition 3.2. Let K Rn a compact set. We define p(K) as ⊆ n p(K) = sup p (0, 1] ; a R with p conv (extK) a K a if it exists 0 { ∈ ∃ ∈ − { − } ⊆ − } otherwise n where extK denotes the set of extreme points of K. Remarks: 1) If p (0, 1] verifies that there exist an a Rn such that p conv (extK) a K a then a∈ K, since 0 is inside the clausure∈ of p conv (extK−) a {, which− is embedded} ⊆ − in K ∈a and so a K. − { − } − ∈ 2) p(K) is an affine invariant of K, i.e. if T = a + S is an affine transformation on Rn with a Rn and S GL(n) then p(T (K)) = p(K). ∈ ∈ 3) The supremum in the last definition can be replaced by maximum, simply by using compactness and continuity arguments. 4)If K is a p-convex body with 0 < p 1 then p(K) p, but if 0 < p < 1 then there are compact sets K with p(K) p which≤ are not p-convex.≥ Notice that p(K) = 1 if and only if K is convex, simply≥ by using Krein-Milman’s theorem. Now we are going to state and prove some upper estimates for the volume ratio of a pair (K1,K2) where K1 is a compact set with vol (K1) > 0 and p(K1) > 0, and K2 is a convex body. We can assume that K1 is in maximal volume position inside K2, since in other case, there would exist an affine transformation T such that T (K1) would be in maximal volume position inside K2 and therefore K1 would work with the pair (T (K1),K2). Hence if p(K1) = p,

1 1 vol (K2) n vol (K2) n vr(K2; K1) = 1 1 vol (K ) n ≤ vol (p conv (extK1) a ) n 1 − { − } 10 for some a K . Therefore ∈ 1 1 1 n vol (K2 a) n vol (conv (extK1) a ) vr(K2; K1) − 1 { − } 1 . ≤ vol (conv (extK ) a ) n vol (p conv (extK ) a ) n { 1 − } − { 1 − }

It can be shown that conv (extK1) a = conv (K1 a) and since conv (K1 a) is in maximal volume position inside{ K − a}we get − − 2 − 1 vol (conv (extK1) a ) n vr(K2; K1) vr(K2; conv (K1)) { − } 1 . ≤ vol (p conv (extK ) a ) n − { 1 − } 1 1 It is easy to check that conv (extK1) a n p − (p conv (extK) a ). Indeed, since a K then { − } ⊆ − { − } ∈ 1

conv (extK1) a = conv (K1 a) = conv x K [0, x a] { − } − {∪ ∈ 1 − } and we can use a stronger version of Caratheodory’s theorem appearing in [E] that asserts that for every x conv (extK1) a there exist xi (extK) a and αi 0, ∈ n { − n} ∈ − ≥ i = 1, . . . , n such that x = i=1 αixi and i=1 αi = 1. Therefore

1 P n p P n p 1 1 α n p − α , i ≤ i i=1 ! i=1 X X 1 1 which implies that x n p − p conv (extK ) a . On the other hand a result of Gi- ∈ − { 1 − } annopoulos, Perissinaki and Tsolomitis (see [G-P-T]) shows that vr(K2; conv (extK1) a ) n and thus we sumarize all these things in the following result { − } ≤ n Proposition 3.3. Let K1,K2 R be such that K1 is a compact set with vol (K1) > 0, ⊆ p(K1) = p > 0 and K2 a convex body. Then

1 vr(K ; K ) n p . 2 1 ≤

Next we are going to prove that if K1 or K2 has some kind of symmetry properties then this general estimate can be slightly improved by using decompositions of the identity in the sense of theorem 1.1, following the spirit of K.M. Ball (see [Ba1]) and A. Giannopoulos, I. Perisinaki, A. Tsolomitis ([G-P-T]). We start with a result which can be found in [G-P-T] and whose proof involves Cauchy-Binet formula. n Lemma 3.4. Let λ1, . . . , λN > 0. Let x1, . . . , xN and y1, . . . , yN be vectors in R N satisfying x , y = 1, for all k = 1,...,N and λ y x = I . Then D D 1, h k ki k k ⊗ k n x y ≥ kX=1 where Dx and Dy are deﬁned by

N det( k=1 λkαkxk xk) Dx = inf ⊗ ; αk > 0, k = 1,...,N (4) N λk ( P k=1 αk ) Q 11 N det( k=1 λkαkyk yk) Dy = inf ⊗ ; αk > 0, k = 1,...,N . (5) N λk ( P k=1 αk ) n Proposition 3.5. Let K1,K2 QR be such that K1 is a symmetric compact set with ⊆ vol (K1) > 0, p(K1) = p > 0 and K2 is a symmetric convex body. Then

1 1 1 vr(K ; K ) n! n n p − . 2 1 ≤

Proof: First of all it is easy to check that we can assume that K1 and K2 are centrally symmetric and so it is ext K1. By using the same arguments than before we conclude that 1/p 1 vr(K ; K ) vr(K ; conv (K ))n − . 2 1 ≤ 2 1 Next we are going to give an upper estimate for vr(K2; L), where K2 and L = conv (K1) are centrally symmetric convex bodies and L is in maximal volume position inside K2. By using theorem 1.1, we can ﬁnd contact pairs (xi, yi) and λi > 0, for all i = 1,...,N , N n2 + n, such that ≤ N N λ y x = I and λ y = 0. k k ⊗ k n k k kX=1 kX=1 n If we take X = conv x1,..., xN L and Y = y R ; y, yk 1 k = 1,...,N K Y , we obtain that{± ± } ⊆ { ∈ |h i| ≤ } 2 ⊆ 1 1 vol (K2) n vol (Y ) n vr(K2; L) = 1 1 . vol (L) n ≤ vol (X) n

Therefore if we ﬁnd some upper estimate for vol (Y ) and lower estimate for vol (X) we will obtain some upper estimates for vr(K2; K1).

n Claim 1: vol (Y ) 2 ≤ √Dy

Consider gj : R R, j = 1,...,N, deﬁned by gj(t) = χ[ 1,1](t). By using the Brascamp-Liev inequality−→ (see [Bar]) we obtain that −

N N λk 1 λk λk 1 1 (gk( x, yk )) dx gk(t) dt = dt Rn h i ≤ Dy R Dy 1 P Z kY=1 kY=1 Z Z− p p where Dy was deﬁned in (5). On the other hand, we conclude that

N λk (gk( x, yk )) dx = χY (x) dx = vol (Y ). Rn h i Rn Z kY=1 Z n Therefore vol (Y ) 2 ≤ √Dy

12 Claim 2: vol (X) 2n √Dx ≥ n! We deﬁne for every x Rn ∈ N N

N(x) = inf αk ; x = αk xk ( | | ) kX=1 kX=1 which is an integrable function that veriﬁes

N N N(x) αp e− dx = sup e− k ; αk 0, x = αkxk dx Rn Rn ( ≥ ) Z Z kY=1 kX=1 N N λk = sup fk(θk) ; x = λkθkxk dx Rn ( ) Z kY=1 kX=1 t where fk : R R is deﬁned by fk(t) = e−| |. Now, if we use the reverse of the Brascamp-Liev−→ inequality (see [Bar]) we can assert that

N N N λk λk sup fk(θk) ; x = λkθkxk dx Dx fk(t) dt n ≥ R (k=1 k=1 ) k=1 R Z Y X p Y Z N (6) λk = Dx 2 k=1 p Y n = Dx2 p where Dx was deﬁned in (4). N(x) On the other hand, we can compute directly the integral of e− by

+ + N(x) ∞ t ∞ t e− dx = e− dt dx = e− dxdt. n n N(x) 0 N(x) t ZR ZR Z Z Z{ ≤ } It is easy to check that x Rn; N(x) t = tX, for all t > 0, and hence { ∈ ≤ } + N(x) ∞ t n e− dx = e− t vol (X) dt = n! vol (X). (7) n ZR Z0 So, combining (6) and (7) we conclude the desired lower estimate for vol (X) and by using Claim 1, Claim 2 and lemma 3.4 we obtain that

1 vr(K ; L) n! n 2 ≤ and hence, the result holds.

By using similar arguments we can prove the following result

13 n Proposition 3.6. Let K1,K2 R are such that K1 is a compact set with vol (K1) > ⊆ 0, p(K1) = p > 0 and K2 is a convex body, then: 1 1 1 e n p (1) If K1 is symmetric, vr(K2; K1) vr(K2; K1) 2 (n!) n − . ≤ ≤ 1 1 1 (2) If K is symmetric,vr(K ; K ) vr(K ; K ) 2 (n!) n n p − . 2 2 1 ≤ 2 1 ≤ Proof: t (1) Takeg ˜j(t) = e χ( ,1](t) instead of gj(t) in the proof of proposition 3.5. ∞ ˜ t (2) Take fj(t) = e− χ[0,+ )(t) instead of fj(t) in the proof of proposition 3.5 and substitute N(x) by ∞

N N ˜ inf αk; αk 0, x = αk xk if it exists N(x) = ≥  (k=1 k=1 )  + X X otherwise. ∞ 

References [Ba1] K.M. Ball, Volume Ratios and a reverse isoperimetric inequality, J. London Math. Soc. (2) 44 (1991), pp. 351-359. [Ba2] K.M. Ball, Ellipsoids of Maximal Volume in convex bodies, Geometria Dedicata 41 (1992), pp. 241-250. [Ba3] K.M. Ball, An Elementary Introduction to Modern Convex Geometry in Flavours of Geometry. Edited by S. Levy. Cambridge University Press. (1997), pp. 1-55. [Bar] F. Barthe, Inégalitésde Brascamp-Liev et convexité, C.R. Acad. Sci. Paris 324 (1997), pp. 885-888. [E] H.G. Eggleston, Convexity, Cambridge Tracts in Math., vol. 47, Cambridge Univ. Press, London and New York (1969). [G-P-T] A. Giannopoulos, I. Perissinaki and A. Tsolomitis, John’s Theorem for an arbitrary pair of convex bodies, preprint. [J] F. John, Extremum problems with inequalities as subsidiary conditions, Courant Anniversary Volume, New York (1948), pp. 187-204 [TJ] N. Tomczak-Jaegermann, Banach-Mazur distances and finite dimensional operator ideals, Pitman Monographs 38 (1989), Pitman, London.

Jesus Bastero: departamento de matematicas,´ universidad de zaragoza, 50009-zaragoza, spain. e-mail:[email protected] Miguel Romance: departamento de matematicas,´ universidad de zaragoza, 50009-zaragoza, spain. e-mail:[email protected]